Question: Simplify and expand the following expression: $ \dfrac{2}{3z + 3}- \dfrac{5}{3z + 21}- \dfrac{4}{z^2 + 8z + 7} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3z + 3} = \dfrac{2}{3(z + 1)}$ We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3z + 21} = \dfrac{5}{3(z + 7)}$ We can factor the quadratic in the third term: $ \dfrac{4}{z^2 + 8z + 7} = \dfrac{4}{(z + 1)(z + 7)}$ Now we have: $ \dfrac{2}{3(z + 1)}- \dfrac{5}{3(z + 7)}- \dfrac{4}{(z + 1)(z + 7)} $ The least common multiple of the denominators is: $ 9(z + 1)(z + 7)$ In order to get the first term over $9(z + 1)(z + 7)$ , multiply by $\dfrac{3(z + 7)}{3(z + 7)}$ $ \dfrac{2}{3(z + 1)} \times \dfrac{3(z + 7)}{3(z + 7)} = \dfrac{6(z + 7)}{9(z + 1)(z + 7)} $ In order to get the second term over $9(z + 1)(z + 7)$ , multiply by $\dfrac{3(z + 1)}{3(z + 1)}$ $ \dfrac{5}{3(z + 7)} \times \dfrac{3(z + 1)}{3(z + 1)} = \dfrac{15(z + 1)}{9(z + 1)(z + 7)} $ In order to get the third term over $9(z + 1)(z + 7)$ , multiply by $\dfrac{9}{9}$ $ \dfrac{4}{(z + 1)(z + 7)} \times \dfrac{9}{9} = \dfrac{36}{9(z + 1)(z + 7)} $ Now we have: $ \dfrac{6(z + 7)}{9(z + 1)(z + 7)} - \dfrac{15(z + 1)}{9(z + 1)(z + 7)} - \dfrac{36}{9(z + 1)(z + 7)} $ $ = \dfrac{ 6(z + 7) - 15(z + 1) - 36} {9(z + 1)(z + 7)} $ Expand: $ = \dfrac{6z + 42 - 15z - 15 - 36}{9z^2 + 72z + 63} $ $ = \dfrac{-9z - 9}{9z^2 + 72z + 63}$ Simplify: $ = \dfrac{-z - 1}{z^2 + 8z + 7}$